A) \[{{A}^{n}}=nA+(n-1)I\]
B) \[{{A}^{n}}={{2}^{n-1}}A+(n-1)I\]
C) \[{{A}^{n}}=nA-(n-1)I\]
D) \[{{A}^{n}}={{2}^{n-1}}A-(n-1)I\]
Correct Answer: C
Solution :
\[{{A}^{2}}=\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]\] \[{{A}^{3}}=\left[ \begin{matrix} 1 & 0 \\ 2 & 1 \\ \end{matrix} \right]\,\left[ \begin{matrix} 1 & 0 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 3 & 1 \\ \end{matrix} \right]\] \ \[{{A}^{n}}=\left[ \begin{matrix} 1 & 0 \\ n & 1 \\ \end{matrix} \right]\] \[nA=\left[ \begin{matrix} n & 0 \\ n & n \\ \end{matrix} \right],(n-1)I=\left[ \begin{matrix} n-1 & 0 \\ 0 & n-1 \\ \end{matrix} \right]\] \[nA-(n-1)I=\left[ \begin{matrix} 1 & 0 \\ n & 1 \\ \end{matrix} \right]={{A}^{n}}\].You need to login to perform this action.
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